Product of Algebraic Expressions – Maharashtra Board Class 7 Solutions for Mathematics

Product of Algebraic Expressions – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise 23:

Solution 1:

Solution 2:

1. 2(3 + 5) = 2 × 3 + 2 × 5
2. 6(7 + 4) = 6 × 7 + 6 × 4
3. 10 (9 – 7) = 10 × 9 – 10 × 7

Solution 3:

The pairs of like terms are:
7y, 2y
5ab, 9ab
11ax², 10x²a

Solution 4:

Solution 5:

Exercise 24:

Solution 1:

Exercise 25:

Solution 1:

Exercise 26:

Solution 1(1):

Solution 1(2):

Solution 1(3):

Solution 1(4):

Solution 1(5):

Exercise 27:

Solution 1(1):

Solution 1(2):

Solution 1(3):

Solution 1(4):

Solution 1(5):

Solution 1(6):

Solution 1(7):

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Solution 1(8):

Solution 1(9):

Solution 1(10):

Solution 1(11):

Solution 1(12):

Exercise 28:

Solution 1(1):

4(y² – 2y + 7)
= (4 × y²) – (4 × 2y) + (4 × 7)
= 4y² – 8y + 28

Solution 1(2):

6(2a² – 3a – 5)
= (6 × 2a²) – (6 × 3a) – (6 × 5)
= 12a² – 18a – 30

Solution 1(3):

7x(3x² – 5x + 8)
= (7x × 3x²) – (7x × 5x) + (7x × 8)
= 21x³ – 35x² + 56x

Solution 1(4):

(m² – 4m + 6)2m
= (m² × 2m) – (4m × 2m) + (6 × 2m)
= 2m³ – 8m² + 12m

Solution 1(5):

(4p² + p + 3)9p
= (4p² × 9p) + (p × 9p) + (3 × 9p)
= 36p³ + 9p² + 27p

Solution 1(6):

3a(a² + 4a – 6)
= (3a × a²) + (3a × 4a) – (3a × 6)
= 3a³ + 12a² – 18a

Solution 1(7):

5q(2a + 3b – 4c)
= (5q × 2a) + (5q × 3b) – (5q × 4c)
= 10aq + 15bq – 20cq

Solution 1(8):

4m(3mn – 2n² + 5)
= (4m × 3mn) – (4m × 2n²) + (4m × 5)
= 12m²n – 8mn² + 20m

Solution 1(9):

(x²y² + xy + 1)xy
= (x²y² × xy) + (xy × xy) + (1× xy)
= x³y³ + x²y² + xy

Solution 1(10):

11p²q(p² – 3pq + 8q²)
= (11p²q × p²) – (11p²q × 3pq) + (11p²q × 8q²)
= 11p⁴q – 33p³q² + 88p²q³

Exercise 29:

Solution 1(1):

(3y – 2)(y² – 5y + 7)
= 3y × (y² – 5y + 7) – 2 × (y² – 5y + 7)
= (3y × y² – 3y × 5y + 3y × 7) – (2 × y² – 2 × 5y + 2 × 7)
= 3y³ – 15y² + 21y – 2y² + 10y – 14
= 3y³ – 17y² + 31y – 14

Solution 1(2):

(a² – ab + b²)(a + b)
= (a² – ab + b²) × a + (a² – ab + b²) × b
= a³ – a²b + ab² + a²b – ab² + b³
= a³ + b³

Solution 1(3):

(2p – 5)(3p² + p – 2)
= 2p × (3p² + p – 2) – 5 × (3p² + p – 2)
= 2p × 3p² + 2p × p – 2p × 2 – 5 × 3p² – 5 × p + 5 × 2
= 6p³ + 2p² – 4p – 15p² – 5p + 10
= 6p³ – 13p² – 9p + 10

Solution 1(4):

(m² – 3mn + n²)(5mn + 4)
= (m² – 3mn + n²) × 5mn + (m² – 3mn + n²) × 4
= m² × 5mn – 3mn × 5mn + n² × 5mn + m² × 4 – 3mn × 4 + n² × 4
= 5m³n – 15m²n² + 5mn³ + 4m² – 12mn + 4n²

Solution 1(5):

(9y – 4x)(8y² – 5xy + 3x²)
= 9y × (8y² – 5xy + 3x²) – 4x × (8y² – 5xy + 3x²)
= 9y × 8y² – 9y × 5xy + 9y × 3x² – 4x × 8y² + 4x × 5xy – 4x × 3x²
= 72y³ – 45xy² + 27x²y – 32xy² + 20x²y – 12x³
= 72y³ – 77xy² + 47x²y – 12x³

Solution 1(6):

(4m² + 6mn + 9n²)(2m – 3n)
= (4m² × 2m + 6mn × 2m + 9n² × 2m) – (4m² × 3n + 6mn × 3n + 9n² × 3n)
= 8m³ + 12m²n + 18mn² – 12m²n – 18mn² – 27n³
= 8m³ – 27n³

Solution 1(7):

(p² + 2q)(7p² – pq – 11)
= p² × (7p² – pq – 11) + 2q × (7p² – pq – 11)
= (p² ×7p² – p² × pq – p² × 11) + (2q × 7p² – 2q × pq – 2q × 11)
= 7p⁴ – p³q – 11p² + 14p²q – 2pq² – 22q

Solution 1(8):

(ab + a²b + ab²)(a + b)
= (ab + a²b + ab²) × a + (ab + a²b + ab²) × b
= (ab × a + a²b × a + ab² × a) + (ab × b + a²b × b + ab²× b)
= a²b + a³b + a²b² + ab² + a²b² + ab³
= a²b + a³b + 2a²b² + ab² + ab³

Solution 1(9):

(x + 2y)(x² – 2xy + 4y²)
= x × (x² – 2xy + 4y²) + 2y × (x² – 2xy + 4y²)
= x × x² – x × 2xy + x × 4y² + 2y × x² – 2y × 2xy + 2y × 4y²
= x³ – 2x²y + 4xy² + 2x²y – 4xy² + 8y³
= x³ + 8y³

Solution 1(10):

(3a – 2b)(9a² + 6ab + 4b²)
= 3a × (9a² + 6ab + 4b²) – 2b × (9a² + 6ab + 4b²)
= 3a × 9a² + 3a × 6ab + 3a × 4b² – 2b × 9a² – 2b × 6ab – 2b × 4b²)
= 27a³ + 18a²b + 12ab² – 18a²b – 12ab² – 8b³
= 27a³ – 8b³